= Solution
\b[Under the ordinary height-and-width definition of an $\aleph_1$-tree, the printed claim is false.] The chain $(\omega_1,\le)$ has one node at each level, every antichain has size at most one, and it is itself an uncountable chain.
Here is the intended argument under the additional splitting convention. Any uncountable chain in a tree with countable levels is unbounded in height, because a bounded set of levels below $\omega_1$ has only countably many nodes. Its predecessor closure therefore gives a <cofinal branch> $B$. At each node on $B$, splitting supplies an extension off $B$, incompatible with a later node on $B$. Recursively for $\xi<\omega_1$, choose such an off-branch node $s_\xi$, then move sufficiently far along $B$ that all subsequent choices are above a branch node incompatible with $s_\xi$. At a limit stage the previous countably many heights are bounded below $\omega_1$, so the recursion continues. The $s_\xi$ are pairwise incompatible, an uncountable antichain.
Thus \b[a splitting $\aleph_1$-tree with only countable antichains has no uncountable chains]. A splitting convention must be stated; the raw tree hypothesis alone does not suffice.
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